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Homework Help: Energy in a spring-mass system

  1. Feb 22, 2009 #1
    1. The problem statement, all variables and given/known data
    A horizontal spring with stiffness 0.6 N/m has a relaxed length of 14 cm (0.14 m). A mass of 24 grams (0.024 kg) is attached and you stretch the spring to a total length of 23 cm (0.23 m). The mass is then released from rest. What is the speed of the mass at the moment when the spring returns to its relaxed length of 14 cm (0.14 m)?

    2. Relevant equations
    Using the Energy Principle:
    Kf + Uf = Ki + Ui + W (Kf and W cancel out in this equation), therefore:
    (1/2) * ks * sf^2 = [ (1/2) * m * vi^2 ] + [ (1/2) * ks * si^2 ]

    3. The attempt at a solution
    I've listed all the variables given:
    ks = 0.6 N/m
    Lo = 0.14 m
    m = 0.024 kg
    si = 0.09 m
    sf = 0 m

    Assuming that I'm looking for the speed of the mass when the spring is at its relaxed length (sf or Lo), I tried plugging in the variables to the energy principle, and got:

    0 = [ (1/2) * 0.024kg * Vi^2 ] + [ (1/2) * 0.6 N/m * (0.09 m)^2 ]

    However, this confuses me as I thought the question is asking for the final velocity, when the spring is at its relaxed position, and when I try solving this equation for Vi I get a negative number..

    Ok so a friend just came by and solved it in 5 minutes, using this equation:
    deltaU -> delta K
    sqrt[ { (1/2) * 0.6 * (0.09^2) } / m ] = v

    Could someone kindly explain the concept to me?

  2. jcsd
  3. Feb 22, 2009 #2


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    You made a few errors. When applying your conservation of energy equations, why did you say W cancels Kf? And why did you use a non zero value for Ki when it starts from rest? Note that "W" in your equation refers to work done by non conservative forces only. Are there any such forces actng?
  4. Feb 22, 2009 #3
    Wow you're fast, thanks for your reply Jay.

    I said W cancels Kf because my book canceled them out, for whichever reason I don't know why. I used a non-zero value for Ki mainly because I didn't really know what I was doing, but I realize now that initial kinetic energy in this case is zero (maybe in other cases too I'm not sure), and potential energy is whatever number in joules, or in other words:

    K + U = constant

    So when kinetic energy is increasing, potential energy is decreasing.

    I'm still working on the rest, but I feel pretty good about finally getting at least a vague idea on the concept of K and U.
  5. Feb 22, 2009 #4


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    OK, but note that K+U remains constant only when non conservative forces such as weight and spring forces are acting. When other forces act, like friction, applied forces, etc., then K + U is not constant , and you mist use the general form of the energy equation which includes the work done by those non conservative forces.
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